Proving Trigonometric Identities

Type the claim, left side ≡ right side, then your working, one line at a time, each line equal to the one above. Every step is checked at 64 angles spread over a whole turn: a tick when the two lines agree at every one where both are defined, a cross with the first angle where they differ, and a warning when a line is undefined where the claim is not. The keys insert the functions, powers and roots.

Into sines and cosines	$\begin{gathered}\tan x=\frac{\sin x}{\cos x}\\\sec x=\frac{1}{\cos x},\quad\csc x=\frac{1}{\sin x}\end{gathered}$
Squares of sine and cosine	$\begin{gathered}\sin^{2}x+\cos^{2}x=1\\ 1+\tan^{2}x=\sec^{2}x\end{gathered}$
1 ± sin x in a denominator	$(1-\sin x)(1+\sin x)=\cos^{2}x$
A difference of two squares	$a^{2}-b^{2}=(a-b)(a+b)$

Worked example — Proving Trigonometric Identities

right side	64 angles checked

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Proving Trigonometric Identities

To prove an identity is to show that two expressions are equal at every angle where both are defined, by a chain of equal expressions from one side to the other.

Where the formula comes from

Start from the more complicated side, usually the one with more terms, fractions or compound and double angles: there is more there to simplify.
$\tan x+\cot x=\cdots=\sec x\csc x$
Rewrite in sines and cosines when tan, sec, csc or cot are mixed: tan x = sin x/cos x and the reciprocals.
$\tan x=\frac{\sin x}{\cos x},\quad\cot x=\frac{\cos x}{\sin x},\quad\sec x=\frac{1}{\cos x},\quad\csc x=\frac{1}{\sin x}$
Use the algebra you already have: a common denominator, a difference of two squares, the conjugate 1 − sin x to make 1 − sin²x.
$a^{2}-b^{2}=(a-b)(a+b),\qquad(1+\sin x)(1-\sin x)=1-\sin^{2}x$
Replace by an identity where one fits: sin²x + cos²x = 1, 1 + tan²x = sec²x, and the compound and double-angle formulae.
$\sin^{2}x+\cos^{2}x=1,\qquad1+\tan^{2}x=\sec^{2}x,\qquad1+\cot^{2}x=\csc^{2}x$
Stop on the other side, written as it is written. Every line equals the one above wherever both are defined, so the first equals the last, provided no line is undefined where the claim is defined. Do not square both sides of the claim: squaring loses the sign, and a false claim can square to a true one.
$f_{0}=f_{1}=f_{2}=\cdots=f_{n}\ \Rightarrow\ f_{0}\equiv f_{n}$

Problems solved in full

Into sines and cosines 3 steps
Prove that tan x + cot x ≡ sec x csc x.

Open this problem in the tool
1. Write the left side in sines and cosines:
  
  $\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
2. Add over the common denominator sin x cos x:
  
  $=\frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x}$
3. The numerator is 1, and the reciprocals finish it:
  
  $=\frac{1}{\sin x\cos x}=\sec x\csc x$
Answer

tan x + cot x ≡ sec x csc x, wherever sin x and cos x are both non-zero. The page agrees at all 64 of its angles.

$\tan x+\cot x\equiv\sec x\csc x$
A difference of two squares 2 steps
Prove that cos⁴x − sin⁴x ≡ cos 2x.

Open this problem in the tool
Show solution
1. cos⁴x − sin⁴x is a difference of two squares, of cos²x and sin²x:
  
  $\cos^{4}x-\sin^{4}x=(\cos^{2}x-\sin^{2}x)(\cos^{2}x+\sin^{2}x)$
2. The second bracket is 1, and the first is the double-angle formula:
  
  $=(\cos^{2}x-\sin^{2}x)\cdot1=\cos2x$
Answer

cos⁴x − sin⁴x ≡ cos 2x for every x.

$\cos^{4}x-\sin^{4}x\equiv\cos2x$
Multiplying by the conjugate 2 steps
Prove that (1 + sin x)/cos x ≡ cos x/(1 − sin x).

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Show solution
1. Multiply top and bottom by 1 − sin x, so the numerator becomes a difference of squares:
  
  $\frac{1+\sin x}{\cos x}=\frac{(1+\sin x)(1-\sin x)}{\cos x\,(1-\sin x)}=\frac{1-\sin^{2}x}{\cos x\,(1-\sin x)}$
2. 1 − sin²x = cos²x, and one cos x cancels:
  
  $=\frac{\cos^{2}x}{\cos x\,(1-\sin x)}=\frac{\cos x}{1-\sin x}$
Answer

(1 + sin x)/cos x ≡ cos x/(1 − sin x), wherever cos x ≠ 0 and sin x ≠ 1.

$\frac{1+\sin x}{\cos x}\equiv\frac{\cos x}{1-\sin x}$
Compound and double angles together 3 steps
Prove that sin 3x ≡ 3 sin x − 4 sin³x.

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Show solution
1. Write 3x = 2x + x and expand with the compound-angle formula:
  
  $\sin3x=\sin(2x+x)=\sin2x\cos x+\cos2x\sin x$
2. Then the double-angle formulae, choosing the form of cos 2x with only sines in it:
  
  $=2\sin x\cos^{2}x+(1-2\sin^{2}x)\sin x$
3. Replace cos²x by 1 − sin²x and collect:
  
  $=2(1-\sin^{2}x)\sin x+\sin x-2\sin^{3}x=3\sin x-4\sin^{3}x$
Answer

sin 3x ≡ 3 sin x − 4 sin³x for every x.

$\sin3x\equiv3\sin x-4\sin^{3}x$
A claim that is false, and where it breaks 3 steps
Is √(1 − cos²x) = sin x an identity? If not, give an angle where it fails and correct it.

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Show solution
1. The first step is a true identity:
  
  $\sqrt{1-\cos^{2}x}=\sqrt{\sin^{2}x}$
2. The square root of a square is the absolute value, not the number itself:
  
  $\sqrt{\sin^{2}x}=|\sin x|$
3. So the claim fails wherever sin x < 0. At the page's first failing angle:
  
  $x=264{.}98^{\circ}:\quad|\sin x|=0{.}9962,\qquad\sin x=-0{.}9962$
Answer

No. It fails throughout 180° < x < 360°, for example at 264.98°, where the sides are 0.9962 and −0.9962. The identity is √(1 − cos²x) = |sin x|.

$\sqrt{1-\cos^{2}x}\equiv|\sin x|$
Prove, then hence solve 3 steps
Prove that (1 − cos 2x)/sin 2x ≡ tan x, and hence solve (1 − cos 2x)/sin 2x = √3 for 0° < x < 360°.
Show solution
1. Use the form of cos 2x that makes 1 − cos 2x a single term, and the double-angle sine:
  
  $\frac{1-\cos2x}{\sin2x}=\frac{1-(1-2\sin^{2}x)}{2\sin x\cos x}=\frac{2\sin^{2}x}{2\sin x\cos x}=\tan x$
2. So the equation is tan x = √3, with tangent's period 180°:
  
  $\tan x=\sqrt{3}\ \Rightarrow\ x=60^{\circ},\quad x=60^{\circ}+180^{\circ}=240^{\circ}$
3. Check that sin 2x ≠ 0 at both, so the left side is defined there:
  
  $\sin120^{\circ}=\tfrac{\sqrt{3}}{2}\ne0,\qquad\sin480^{\circ}=\tfrac{\sqrt{3}}{2}\ne0$
Answer

x = 60° or x = 240°.

$x=60^{\circ},\ 240^{\circ}$
Find the mistake 2 steps
A student writes cos⁴x − sin⁴x = (cos²x − sin²x)² = cos 2x. Find the mistake.

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Show solution
1. The first step factorises a difference of two squares as a square of the difference:
  
  $\cos^{4}x-\sin^{4}x\ \ne\ (\cos^{2}x-\sin^{2}x)^{2}$
2. The two are not equal. At 42.49°, the page's first failing angle:
  
  $x=42{.}49^{\circ}:\quad\cos^{4}x-\sin^{4}x=0{.}0874,\qquad(\cos^{2}x-\sin^{2}x)^{2}=0{.}0076$
Answer

Step 1 is wrong: a² − b² = (a − b)(a + b), not (a − b)². Its right side is cos²2x, so the working proved nothing, although the claim itself is true.

$a^{2}-b^{2}=(a-b)(a+b)\ne(a-b)^{2}$

Supplementary problems, answers only

Prove that 1/(1 − sin x) + 1/(1 + sin x) ≡ 2 sec²x.

Show answer

Common denominator 1 − sin²x = cos²x; the numerators add to 2.

$\frac{1}{1-\sin x}+\frac{1}{1+\sin x}=\frac{2}{1-\sin^{2}x}=\frac{2}{\cos^{2}x}=2\sec^{2}x$
Prove that (sin x + cos x)² ≡ 1 + sin 2x.

Show answer

Expand: sin²x + cos²x + 2 sin x cos x = 1 + sin 2x.

$(\sin x+\cos x)^{2}=1+2\sin x\cos x=1+\sin2x$
Prove that sec⁴x − tan⁴x ≡ sec²x + tan²x.

Show answer

Factorise as (sec²x − tan²x)(sec²x + tan²x), and sec²x − tan²x = 1.

$\sec^{4}x-\tan^{4}x=(\sec^{2}x-\tan^{2}x)(\sec^{2}x+\tan^{2}x)=\sec^{2}x+\tan^{2}x$
Prove that cot x − tan x ≡ 2 cot 2x.

Show answer

Over sin x cos x the numerator is cos²x − sin²x = cos 2x, and sin x cos x = ½ sin 2x.

$\cot x-\tan x=\frac{\cos2x}{\tfrac12\sin2x}=2\cot2x$
Is sin x/(1 + cos x) ≡ (1 − cos x)/sin x?

Show answer

Yes, wherever both sides are defined: multiply the right side top and bottom by 1 + cos x.

$\frac{1-\cos x}{\sin x}=\frac{1-\cos^{2}x}{\sin x\,(1+\cos x)}=\frac{\sin x}{1+\cos x}$
Is cos x = √(1 − sin²x) an identity?

Show answer

No: it fails wherever cos x < 0, for example at 120°, where the left side is −1/2 and the right side 1/2.

$x=120^{\circ}:\quad\cos x=-\tfrac12,\quad\sqrt{1-\sin^{2}x}=\tfrac12$

Agreement at a handful of angles is evidence, not proof

Press √(1 − cos²x) = sin x? and try it at 30°, 45°, 60°, 90° or 150°, the angles a calculator check reaches for, and the two sides agree every time. The claim is false. The page's first failing angle is 264.98°, where the left side is 0.9962 and the right side −0.9962: a square root is never negative, while sin x is negative all through quadrants III and IV. Checking the half-turn where sin x is positive would have found nothing, which is why the page spreads its 64 angles over the whole turn. What is true is √(1 − cos²x) = |sin x|.

Squaring both sides makes a false identity look proved

The same false claim survives the one move that cannot be undone. Square both sides of √(1 − cos²x) = sin x and you get 1 − cos²x = sin²x, which is true at every angle. Squaring cannot tell sin x from −sin x, so a true statement at the end of a chain of squarings says nothing about the start: −1 = 1 squares to 1 = 1. This is why a "show that" runs one way, from one side to the other, with every line equal to the one above. Each = is then a fact about one and the same number, and the facts chain from the left side to the right.

With whole multiples of x, enough angles are a proof

There is a case where checking angles proves the claim. A sum of sines and cosines of x, 2x, …, nx (a trigonometric polynomial of degree n) that is not zero everywhere is zero at no more than 2n angles in a turn: writing z for the point at angle x on the unit circle turns it into a polynomial in z of degree 2n divided by zⁿ, and such a polynomial has at most 2n roots. So two such expressions that agree at 2n + 1 different angles in a turn agree everywhere, and fractions come down to the same case once the denominators are multiplied out. Agreement at all 64 of the page's angles therefore settles any claim of degree up to 31 after clearing denominators, apart from rounding. Square roots and absolute values fall outside the argument, and that is exactly where the first block's false claim lives.

Learning path

Trigonometric identities and equations

References (1)

The method the lesson states: work on one side, usually the more complex one; factor, square a binomial or add fractions; failing that, convert to sines and cosines: J. Abramson, Precalculus 2e, OpenStax (2021), section 7.1, “Simplifying and Verifying Trigonometric Identities”.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.

Proving Trigonometric Identities

Where the formula comes from

Supplementary problems, answers only

Agreement at a handful of angles is evidence, not proof 🖖

Squaring both sides makes a false identity look proved 🖖

With whole multiples of x, enough angles are a proof 🖖

Agreement at a handful of angles is evidence, not proof

Squaring both sides makes a false identity look proved

With whole multiples of x, enough angles are a proof